laplacetransform te^{2t}cos(5t)

解答

拉普拉斯变换

t e^{2 t} cos (5 t)

- \frac{- s ^{2} + 4 s + 21}{( s ^{2} - 4 s + 29 ) ^{2}}

求解步骤

L {e^{2 t} t cos (5 t)}

使用拉普拉斯变换表:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

对于

t e^{2 t} cos (5 t) : f (t) = e^{2 t} cos (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{2 t} cos (5 t)})

L {e^{2 t} cos (5 t)} : \frac{s - 2}{( s - 2 ) ^{2} + 25}

\frac{d}{d s} (\frac{s - 2}{( s - 2 ) ^{2} + 25}) = \frac{- s ^{2} + 4 s + 21}{( s ^{2} - 4 s + 29 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} + 4 s + 21}{( s ^{2} - 4 s + 29 ) ^{2}}

= - \frac{- s ^{2} + 4 s + 21}{( s ^{2} - 4 s + 29 ) ^{2}}

d er i v a t i v e (10 + sec (θ)) sin (θ)

t a y l or \frac{1}{1 - x}, 8

\frac{d}{d x} (x^{2} ln (x^{5}))

f (x) = 6 e^{x}

\int \frac{5 x ^{2} - 6}{x ^{2} - 2 x - 8} d x